Step 1: Find the principal and simple interest rate. Amount after 3 years: \[ A_1 = 13920. \] Amount after 6.5 years: \[ A_2 = 18960. \] Time difference: \[ 6.5 - 3 = 3.5 \text{ years}. \] Extra interest in these 3.5 years: \[ \Delta I = A_2 - A_1 = 18960 - 13920 = 5040. \] So simple interest per year: \[ \text{SI per year} = \frac{5040}{3.5} = 1440. \] Interest for first 3 years: \[ \text{SI}_{3\text{ yrs}} = 1440 \times 3 = 4320. \] Hence principal: \[ P = A_1 - \text{SI}_{3\text{ yrs}} = 13920 - 4320 = 9600. \] Rate of interest: \[ R = \frac{\text{SI per year}}{P} \times 100 = \frac{1440}{9600} \times 100 = 15% \text{ p.a.} \]
Step 2: Compound interest for 2 years, half-yearly. Principal: \[ P = 9600. \] Rate per half-year: \(r = \frac{15\%}{2} = 7.5% = 0.075\). Number of half-yearly periods in 2 years: \[ n = 2 \times 2 = 4. \] Amount: \[ A = P\left(1 + \frac{r}{100}\right)^n = 9600 \times (1.075)^4. \] Approximate: \[ (1.075)^2 \approx 1.1556,\quad (1.1556)^2 \approx 1.335\ (\text{approximately}), \] so \[ A \approx 9600 \times 1.335 \approx 12820. \] Compound interest: \[ \text{CI} = A - P \approx 12820 - 9600 = 3220 \text{ (approximately)}. \] Among the given options, the nearest value is \(3221\). Therefore, the total interest would have been approximately: \[ \boxed{3221}. \]
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: